If $S=[0,1]$ then show infimum for $S$ is $0$ and Supremum is $1$

Question

These kinds of questions really make me say" its so obvious, why do I even need to prove it" but in real analysis , even $1>0$ needs a prove. So I encountered this question in basic studies of real analysis and I haven't yet taken the course( just preparing) but I already know the definitions of Sup and Inf for sets.Now i tried looking up at how to prove SupS or Inf S, found some equivalent statements related to the definitions of Sup only so i tried using them. Of course starting my proof with the very"obvious. $1$ is the upper bound of the set S, so for all s in S, $s\leq1$ is always true, now I just need to prove its the least upper bound. By contradiction, I let $v<1$ be an upper bound then there exist a number $\frac{(v+1)}{2}$ that belongs in the set and is bigger than v. Therefore i reach a contradiction, v is not an upper bound hence $1$ is the least upper bound. Now I don't know if this prove is complete or if I am missing some points? I need a bit of clarification on the steps i used.

Answer 1

That looks ok. But I guess you'd have to prove $(v+1)/2$ satisfies your claim too. Here's an easier more direct proof. Let $B$ be an upperbound of $S$. Because $1\in S$ and by the definition of upperbound, $1 \le B$. Moreover, by the defn of $S$, any $x \in S$ is $\le 1$. So $1$ satisfies the defn of supremum. The proof for infimum is similar.

Answer 2

Let's follow your proof line by line.

• "$1$ is the upper bound of the set $S$." I would say $1$ is an upper bound of $S$ (there are infinitely many).

• "So for all $s \in S$, $s \leq 1$ is always true." Sure, that's the definition of $S$.

• "Now I need to prove it's the least upper bound." Good, yes.

• "By contradiction, I let $v < 1$ be an upper bound. Then there exists a number $(v+1)/2$ that belongs in the set and is bigger than $v$." Yes. Maybe you should show why $(v+1)/2 > v$. This is equivalent to $v+1 > 2v$ and this is equivalent to $1 > v$, which we assumed.

• "Therefore, I reach a contradiction: $v$ is not an upper bound." Yep, if $v < 1$ then $v$ is not an upper bound. So all upper bounds must satisfy $v \geq 1$. Let $s = \sup(S)$. Then $s$ is the least upper bound, so $s \leq 1$ (by minimality). Since $s$ is an upper bound, we've shown we must have $s \geq 1$. Thus $1 \leq s \leq 1$ which forces $s = 1$.

Here's maybe a faster way to get things. You take $v < 1$. $1 \in S$, so $v$ cannot be an upper bound. Now you don't need to deal with things.

Good job. Seems fine to me.

Answer 3

In this case it's even simpler, since the set has what's known as a maximum and minimum, that is, upper/lower bounds that belong to the set.

When such elements exist on a set $S$, it's clear that they must be the supremum/infimum. For example, the maximum (let's denote it as $M$) is also supremum because any element lower than $M$, say $a$, can't be an upper bound of $S$. Why? Because $a<M$, and $M \in S$.

In your example, the interval $[0,1]$ is precisely defined as the set of points between $0$ and $1$, both included, so these two elements are minimum and maximum, and hence infimum and supremum, respectively.